How do you solve the quadratic using the quadratic formula given 3x^2-11x=8-14x over the set of complex numbers?

1 Answer
Aug 30, 2016

Solution is -1/2-sqrt105/6 and -1/2+sqrt105/6

Explanation:

For the equation ax^2+bx+c=0, the roots are given by x=(-b+-sqrt(b^2-4ac))/(2a). As can be seen type of roots depend upon discriminant b^2-4ac. If it is negative, roots are complex and if it is a complete square (and coefficients are rational), roots are rational.

In the equation 3x^2-11x=8-14x or 3x^2+3x-8=0, the discriminant is (3)^2-4×3×(-8)=9+96=105, which is not a complete square and hence roots are irrational (not complex as mentioned in question).

Roots are x=(-3+-sqrt105)/(2×3)

= (-3+-sqrt105)/6

= -1/2+-sqrt105/6

Hence solution is -1/2-sqrt105/6 and -1/2+sqrt105/6.